Let $\mu$ be the counting measure on the measurable space $(N,P(N))$.
$1 <p,q< \infty$ , $\frac{1}{p}+\frac{1}{q}=1$.
Define: $\phi: l_q\to (l_p)^*$ such as for all $(a_n)\in l_p$ and $(b_n)\in l_q$:
$\phi((b_n)_n)((a_n)_n)=\sum_{n=1}^{\infty} a_n b_n$.
A.Show that $\phi$ is well-defined, linear, injective and $\|\phi(b_n)_n\|\leq \|(b_n)_n\|_q$.
B.Show that $\phi$ is surjective and $\|\phi(b_n)_n\|=\|(b_n)_n\|_q$.
Hints for part B:
1.Let $f\in (l_p)^*$, $\forall n\in N$, denote $e_n(k)$ as:
$e_n(k)$ = 1 if n=k and 0 otherwise.
So $e_n\in l_p$ (why?). And look at $b_n:=f(e_n)$.
- $\forall n\in N$ define:
$\lambda_n= |b_n|^q/ b_n$ if $b_n \neq 0$ and 0 otherwise.
Let $N_0\in N$ show that $\sum_{n=1}^{N_0} (|b_n|^q)^{1/q} \leq \|f\|$.
This what I did (according to my understanding of this question).
$\phi$ is well defined:
It means to make sure that the given sum is in $(l_p)^*$ which is equivalent to say that the defined sum is a function from $l_p$ to $C$.
$|\phi(b_n)(a_n)|=\left|\sum_n a_n b_n\right|\leq \sum_n |a_n b_n| \leq \|a_n\|_p \|b_n\|_q < \infty$, by using the Holder inequality.
$\phi$ is Linear:
Let $b_n , c_n \in l_q$ , $a_n \in l_p$ , $b,c \in C$.
Then, $\phi(b(b_n)+c(c_n))(a_n)= \sum_n (a_n)(b(b_n)+c(c_n))=\sum_n (b(a_n)(b_n)+c(a_n)(c_n))=b\sum_n (a_n)(b_n)+ c\sum_n (a_n)(c_n)= b \phi(b_n)(a_n)+ c \phi (c_n)(a_n)$.
$\phi$ is injective:
Let $(a_n)\in l_p$ , $(b_n), (c_n) \in l_q$.
Assume that : $\phi(b_n)(a_n)=\sum_n (a_n)(b_n)=\phi(c_n)(a_n)=\sum_n (a_n)(c_n)$, so $\sum_n (a_n)((b_n)-(c_n))=0$ and since it is true for every $a_n \in l_p$, we have $b_n=c_n$ ,$\forall n$ thus $(c_n)_n=(b_n)_n$.
And the last thing, $\|\phi(b_n)\| =\|\sum_n (a_n)(b_n)\| \leq \sum_n \|(a_n)(b_n)\| \leq \|a_n\|_p \|b_n\|_q$ ,here i used the holder inequality, so $\|\phi(b_n)\|=\sup_{\|a_n\|_p=1} |\phi(b_n)| \leq \|b_n\|$ .
Part B:
$\phi$ is surjective:
Is is easy to see that $e_n \in l_p$ since
$\|e_n\|_p=(\sum_n |e_n|^p)^{1/p}=1 < \infty$ so $e_n \in l_p$.
(Here we use the p-norm definition), we'll need this later!.
We aim to find a sequence $(b_n)_n\in l_q$ such as $\phi((b_n)_n)=f$. Look at:
$b_n=f(e_n)$ so we get $\phi((b_n)_n)(e_n)=\sum_n e_n*b_n=b_k=f(e_k)=f(e_n)$ (using $e_n$ definition).
We want to show that $b_n\in l_q$:
$\|b_n\|_q= \|f(e_n)\|\leq \|f\|_{l_p^*} \|e_n\|_p = \|f\|_{l_p^*}<\infty$ since we assume that $f\in l_p^*$.
Is what I did fine?
Can you please help me with the other part in 2?
Best Answer
As for the part A, for $\| \phi(b_n)\|$ what is meant is the norm in $\ell_p^*$, the standard dual norm in a Banach space, so, if you apply the definition of dual norm, with your computation you've almost finished.
As for the party B, I didnt understand what you write. The idea of the proof is that you want to prove that there exists $\mathbf{b}=(b_n)_n$ such that $f=\phi(\mathbf{b})$, but of such a sequence exists, it is determined by the fact that $\phi(\mathbf{b})(e_k)= b_k$, so it must necessarily be $f(e_k)= b_k$. Then you have to prove that $\mathbf{b}\in\ell_q$ and that $\phi(\mathbf{b})= f$. The latter is easier, the former a bit trickier